vedanta Excel In Opt. Mathematics - Book 10
is called combined/composition of transformations. Now, let us answer the question.
-What is the combined transformation of above transformation ?
Rotation through 180° about centre origin O is the single transformation of above combined
transformations (reflections) .
Definition
When an object has been transformed, its image can again be transformed to form a new
image. Such transformation is called combination of transformation (or composition of
transformations). After a combination of transformations, the change from the single object
to the final image can be described by a single transformation.
Let R1 be a transformation which transforms a point P to the point P'. Again let R2 be
another transformation which transforms P' to the point P''. Then, the transformation which
transforms the point P to P'' is said to be the composition of R1 and R2. It is denoted by R2oR1
(read as R1 is followed by R2).
R2oR1 means that the first transformation is R1 and it is followed by R2. It means the first
transformation is due to R1 and , it is followed by R2.
R1oR2 means that the first transformation is R2, and it is followed by R1.
Hence, R1oR2 and R2oR1 have different meanings. There may be a single transformation
which represents the combination of given transformations.
11.2 Combination of Translation and Translation
( ) ( )2 4
Let P(1, 3) be a point and T1 = 5 and T2 = –4 be two translation vectors. Firstly
P is translated by T1. We get,
Y
( )P (1, 3)T1 =2 P' (3, 8)
5
P' (1 + 2, 3 + 5) = P' (3, 8)
Again, P' (3, 8) is translated by T2 we get, P"(7, 4)
X
( )P'(3, 8)T2 =4P'' (3 + 4, 8 – 4) = P'' (7, 4) P(1, 3)
-4 O
Y'
Also a single translation that represents the above two X'
translations is given by
T1 T2
2 46 P"
( ) ( ) ( )T = 5 + –4 = 1
( )A (1, 3)
T= 6 P'' (1 + 6, 3 + 1) = P''(7, 4) T2 n
1 P' m C
A translation followed by another translation: T2oT1
T1
( )a b
Let, T1 = b be a translation vector. P aA B
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( )m
T2 = n be another translation vector.
Let P(x, y) be any point.
T1 translates P to P' with 'a' units along the x-axis and 'b' units along the y - axis.
P(x, y) ( )T2 =a P'(x + a, y + b)
b
Again, T2 translates P' to P'' with 'm' units along the x-axis and 'n' units along the y-axis.
P'(x + a, y + b) ( )m P'' (x + a + m, y + b + n)
T2 = n
Now, let us find a single translation vector T which translates P to P''.
Now, x-component of T = PB = PA + AB = PA + P'C = a + m
y - component of T = BP'' = BC + CP'' = AP' + CP'' = b + n
( )a + m
T = T2 o T1 = b+ n
a m
n , the combined
( ) ( )Hence, if translation T1 = b is followed by translation T2 =
a+m
( )translation is given by T = T1 + T2 = b+ n .
Note :
( ) ( )If T1 =
am
b and T2 = n are two translation vectors, the combined translation
a+m
( )T2oT1 have the same effect i.e. T1oT2 = T2oT1 = b + n
Worked out Examples
( ) ( )1
Let T1 = 2
Example 1. –3
Solution: and T2 = 2 be two translation vectors. Then find the
image of point P(2, 3) under the combined translation T2 o T1
302
Given translation vectors are :
( ) ( )1 -3 Y
T1 = 2 and T2 = 2 P"(6, 10)
The combined translation vector of
T1 and T2 is given are 2 T2 P'(3, 5)
-3
( ) ( ) ( )T = T1 + T2 = T1 1
1 -3 -2
2 + 2 = 4 P(2, 3)
( )-2 X' O X
Now, P(2, 3) T = 4 P' (2 - 2, 3 + 4)
= O' (0,7) Y'
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Alternative Method
T2 o T1 means translation T1 is followed by T2
( )P(2, 3)T= 1 P' (2 + 1, 3 +2) = P'(3, 5)
2
( )-3 P' (3 - 3, 5 + 2) = P'(0, 7)
Again, P'(3, 5) T = 2
( ) ( )Let T1 =2 4
Example 2. 4 and T2 = 5 be two translations vectors, what point
Solution:
would have the image (9, 12) under the combined translation T2 o T1 ?
( ) ( )Here, T1 =2 4 , combined translation
4 T2 = 5
( ) ( ) ( )vector T of T2 o T1 = 4 + 5 i.e. T = 9
24 6
Let P(x, y) be the required point.
( )Then, P(x, y)T= 6 P' (x + 6, y + 9) But P'(9, 12)
9
? (9, 12) = ( x + 6, y + 9)
Then equating the corresponding elements, we get,
9 =x + 6 or, x = 3
and 12 = y + 9 or, y = 3
? P(x, y) = P(3, 3)
Example 3. ( ) ( )-1 2
Solution:
Let T1 = 3 and T2 = 3 be two translation vectors. Find the image
of the vertices of ∆ABC with A(2, 3), B(4, 5) and C(6, 3), under composite
translation T2 o T1. Show ∆ABC and ∆A'B'C' on the same graph.
-1( ) ( )Here, T1 = 2 Y
3 and T2 = 3 B'
Combined translation vector A' C'
( ) ( )-1 2 B
T = T1 + T2 = 3 + 3
= ( 1 ) A C
6 X
O
( )1 X' Y'
Under translation T2 o T1 = T = 6
( )A(2, 3)
T= 1 A' (2 + 1, 3 + 6) = A1 (3, 9)
6
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B(4, 5) T = (61 ) B' (4 = 1, 5 + 6) = B'(5, 11)
C(6, 3) T = (16 ) C'(6 + 1, 3 + 6) = C'(7, 9)
The graphs of ∆ ABC and ∆A'B'C' are plotted alongside.
Exercise 11.1
Very Short Question
1. (a) Define combined transformation.
(b) If T1 and T2 are two transformations, write a difference between T1oT2 and T2 o T1.
( ) ( )(c) Write the combination of two translations T1 = a m
b and T2 = n .
Short Questions
( ) ( )2 -5
2. If T1 = 1 and T2 = 6 are two translations, find the images of given points
under the given combined translations.
(a) T2 o T1 (2, 5) (b) T1 o T2 (6, 5)
(c) T12 (2, 4) (d) T22 (3, 5)
( ) ( )3. 3 1
Let T1 = 2 and T2 = 2 be two translations, find the image of the following
points under the combined translations T1 o T2 and T2 o T1
(a) A (4, 5), (b) B(-6, 7) (c) C(2, -5) (d) D(2, 3)
( )4. a
(a) The image of a point (3, 4) is the point (10, 10) under a translation b followed
( )by another translation 4 . Find the values of a and b.
5
( ) ( )2
(b) Let T1 = 3
4
and T2 = 4 be two translation vectors. What point would
have the image (-8, 12) under the combined translation T2 o T1.
( ) ( )(c) If T1 = 1 -2 and T1 o T2 (x, y) = (8, 8) find the values of
x and y. 2 and T2 = 4
( ) ( )(d) If P =4 -4 and ToP (x, y) = (6, -6), find the values of x and y.
5 , T= 2
(e) If T1 and T2 are two translations defined by T1(x, y) = (x + 6, y - 3), and
T2(x, y) = (x - 3, y + 4). Find the image of P(1, 3) under T2 o T1 .
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Long Questions
( ) ( )-1 -3
5. Let T1 = 2 and T2 = 4 be two translations. Translate ∆ABC with vertices
A(2, 2), B(-2, 2), and C(6, 6) under translation T1 and then T2.
(a) Find the image ∆A'B'C' of ∆ABC.
(b) Find the image ∆ A" B" C" of ∆ A'B'C' under translation T2.
(c) Draw graphs of ∆ABC, ∆A'B'C' and ∆A"B"C" on the same graph paper.
( ) ( )3 1
6. Let T1 = -2 and T2 = 1 be two translation vectors. If P(2, 3), Q(4, 5), R(6, 4),
S(7, 8) are the vertices of quadrilateral PQRS. Find the image of quadrilateral PQRS
due to transformation T2 o T1 . Show the quadrilateral PQRS and its image on the same
graph.
7. P(1, 2), Q(7, 2), R(7, 9) and S(1, 9) are the vertices of rectangle PQRS. Draw it on a graph
paper . Find the coordinates of the vertices of the image rectangle P'Q'R'S' under the
( ) ( )2 2
transformation T1 o T2 if T1 = 3 and T2 = -2 . Also draw the image on the same
graph paper.
8. If T1 and T2 be two translations defined by T1(x, y) = (x + 3, y + 2) and T2(x, y) =(x - 4,
y - 7). Find T1 o T2 (x, y) and T2 o T1, (x, y). Also find the images of A(1, 2), B(7, 2) and
C(4, 7) of ∆ABC under the composite translation T1 o T2 and T2 o T1. Are images due to
T1oT2 and T2oT1 same ? Show ∆ABC and it images on the ame graph paper.
( )1.
(c) a+m 2.(a) (-1, 12) (b) (3, 12) (c) (6, 6)
b+n 3.(a) (8, 9) (b) (-2, 11) (c) (6, -1)
(d) (-7, 17)
(d) (6, 7) 4.(a) a = 3, b = 1 (b) (-14, 5) (c) x = 9, y = 2
(d) x = 6, y = -13 (e) (4, 4)
5. A'(1, 4), B'(-3, 4), C'(5, 8), A"(-2, 8), B"(-6, 8), C"(2, 12)
6. P'(6, 2), Q'(8, 4), R'(10, 3), S'(11, 7)
7. P'(5, 3), Q'(11, 3), R'(11, 10), S'(5, 10) 8. A'(0, -3), B'(6, -3), C'(3, 2)
11.3 Combinations of Rotation and Rotation
A rotation followed by another rotation can be expressed as a single rotation. We have the
following two cases:
A. When two rotations have same centre.
A point or an object once rotated about a centre through a given angle can further be
rotated about the same centre through another given angle.
Let P(x, y) be any given point in the plane.
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Let it be rotated through positive 90° about origin.
P(x, y) P'(-y, x)
Again, let P' (-y, x) be further rotated through positive 90° about the same centre O.
P (- y, x) P' (-x, -y)
Here the final image of P is P".
But P" (- x, - y) is the image of the point P(x, y) under the rotation of 180° about the
origin.
Y
P'(–y, x)
P(x, y)
Dq Eq Dq
X' Eq X
O
P"(–x, –y)
Y'
Hence, a rotation of D° followed by another rotation of E° about the same centre is
equivalent to the rotation of (D° + E°) about the same centre.
Worked Out Examples
Example 1. Let P(4, 2), Q(6, 2) and R(5, -1) be the vertices of ∆PQR. Find the coordinates
Solution: of image ∆P'Q'R' under the rotation through 90° followed by a rotation
through 180° about the origin.
The first angle of rotation (D°) = 90° about origin O.
The second angle rotation (E°) = 180° about origin O.
Since the centre of rotation in both cases is same O.
Hence, combined rotation can be expressed as a single rotation (90° + 180°)
= 270° positive about the same centre O.
The vertices of image ∆ P'Q'R' due to combined rotations are as follows :
P(x, y) R[(0, 0), 270°] P'(y, –x)
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P(4, 2) P'(2, -4)
Q(6, 2) Q'(2, -6)
R(5, -1) R' (-1, -5)
Example 2. A(2, 3), B(4, 5) and C(5, 1) are the vertices of a triangle ABC. The triangle ABC
Solution: is first rotated about the origin through positive 90°. Find the coordinates of
vertices image ∆A'B'C'. Again ∆A'B'C' is rotated through 90° in anticlock. Find
the coordinates of image ∆A"B"C". Plot ∆ABC, ∆A"B"C" on the same graph,
∆A"B"C". what is the single transformation of the combined rotation.
The first angle of rotation = 90° about centre O.
The images due to rotation of +90° about centre O of A, B, C, are given below:
P(x, y) R[(0, 0), +90°] (–y, x) Y
A(2, 3) R[(0, 0), +90°] A'(-3, 2) C' B
B(4, 5) R[(0, 0), +90°] B'(-5, 4) B'
A
A'
C(5, 1) R[(0, 0), +90°] C'(-1, 5) C
X' X
C" O
Again, ∆A'B'C' is rotated through
+90° about the same centre O. A"
Its image vertices are given below: B"
A'(-3,2) R[(0, 0), +90°] A" (-2, -3) Y'
B'(-5, 4) R[(0, 0), +90°] B" (-4, -5)
C'(-1, 5) R[(0, 0), +90°] C" (-5, -1)
Graphs of ∆ABC, ∆A'B'C' and ∆A"B"C"are plotted on the graph.
Since, the centre of rotation is same in given two rotations, the combined
rotation is (90° + 90°) = 180° about the same centre O in positive 180°.
B. When two rotations are about the different centres.
A point or an object once rotated about a centre through a given angle can further be rotated
Y
about the different centre through another given angle.
Let R1 be the rotation through an angle x° about a centre A C
and R2 be another rotation through an angle y° about the C' B'
different centre. Then, the equivalent transformation B
(R2oR1) is again a new rotation through the angle (x°+y°) A' RP X
about the third centre. This centre is the meeting point of O
the perpendicular bisectors of the line segments joining X' B"
corresponding vertices of the object and its final image. In Q
other words, perpendicular bisectors of the line segments
joining corresponding vertices of the object and its final A"
image are concurrent and the point of concurrence is the C"
centre of new rotation (i.e., combined rotation).
Y'
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Let ABC be the given triangle (object). It is rotated to ∆ A'B'C' through 90° about the centre
P. The triangle A'B'C' is again rotated to ∆ A"B"C" through 90° about the centre Q. Then
∆A"B"C" is the image of ∆ ABC under the rotation through 180° about the third centre R.
Perpendicular bisectors of AA", BB", and CC" intersect each other at the point R.
Example 3. Let R1 = Rotation of 90° about origin in anti clockwise and R2 = rotation
Solution: of 180° in anti clockwise about centre (2, 3). Then, find the image of
P(5, 3) under combined transformation R2 o R1 (P).
Here, R1 = rotation of +90° about O.
R2 ritation of 180° about (2, 3)
Since (x, y) R[(0, 0), +90°] (-y, x)
and (x, y) R[(a, b), 180°] (2a - x, 2b - y)
Now, P (5, 3) P'(-3, 5)
Again P'(-3, 5) R[(a, b), 180°] P" (2.2 + 3,2.3 - 5) = P" (7, 1)
Combined transformation of Rotation and Translation.
Let P(x, y) be any point on the plane.
( )Let R = rotation of + 90° about origin and T = translation vector = a
b
We find the image due to ToR of P, under rotation of + 90° about O.
p(x, y) R[(0, 0), 90°] p' (-y, x)
( )Again, P'(-y, x) T= a P' (-y + a, x + b)
b
Similar relations can be established for RoT and other rotation followed by
translation of translation followed by rotation.
Example 4. ( )A(2, 1), B(7, 2) and C(3, 6) are the vertices of ∆ABC. Let T = 3 and
Solution: 2
R = rotation of + 90° about origin. Find the image of ∆ABC under RoT. Draw
graphs of ∆ABC and its image in the same graph paper.
( )Here, T = 3 , R = rotation of +90° about O. ∆ABC is translated by T, we
2
get the image ∆A'B'C'.
T=( )P(x, y) 3 (x + a, y + b)
2
T=( )A(2,1) 3 A'(2 + 3, 1 + 2) = A' (5, 3)
2
( )B(7, 2)T= 3 B'(7 + 3, 1 + 2) = B'(10, 4)
2
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( )C(3, 6) T= 3 C'(3 + 3, 6 + 2) = C' (6, 8)
2
Again, ∆A'B'C' is rotated about O B" Y
through +90°. C
C'
A'(5, 3) A"(-3, 5) C" A"
B'(10, 4) B"(-4, 10)
A' B B'
X
C'(6, 8) C"(-8, 6) X' A
O
Graphs of ∆ABC, ∆A'B'C' and Y'
∆A"B"C" are plotted.
Exercise 11.2
Short Questions
1. Let P(2, 3) be a point in plane and R1 = rotation of +90° about origin.
R2 = rotation of 180° about origin.
(a) Find the image of the point P under the following transformations:
(i) R1oR2 (ii) R2oR1
(b) Write down a single transformation of above combined.
2. Let, R1 = rotation of +90° about origin.
R2 = rotation of +90° about origin.
(a) Find the image of the following points under the following transformations:
(i) R1 o R2 (4, 5) (ii) R1 o R2 (-3, -2) (iii) R2 o R1(4, 6)
(b) Write down a single transformation of combined transformation.
(i) R2oR1 and (ii) R1 o R2
3. (a) Find the image of P(2, 3) due to combined transformation.
(i) R1oR2 where R1 = rotation of +90° about origin.
(ii) R2R1 R2 = rotation of + 90° about (2, 5)
(b) Find the image of M(2, 5) due to combined transformation
(i) R1oR2 where R1 = rotation of +90° about (1, 2)
R2 = rotation of 180° about O
(ii) R2 o R1
(c) Find the image of S(7, 5) under combined transformation R2 o R1, where,
R1 = rotation of 180° about (2, 3)
R2 = rotation of +90° about origin.
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( )4. (a) Let T =2 and R = rotation of +90° about O.
3
Then find the images of given points under.
(i) ToR (2, 7) (ii) RoT (4, 5) (ii) ToR (-2, -3)
( )(b) Let R = rotation of 180° about origin. T = -2
3
Find the images of A(2, 3) and B(6, 7) under (i) ToR (ii) RoT
Long Questions
5. A(2, 5), B(6, 5) and C(2, -3) are vertices of ∆ABC. Rotate ∆ABC through +90° about
origin and obtain image ∆A'B'C'. Again rotate ∆A'B'C' about the same centre origin
through 180° obtain images ∆A"B"C", plot the triangle ABC and its images on the same
graph paper. What is a single transformation for above transformation?
6. Points A(5, 2), B(4, 5) and C(8, 4) are the vertices ∆ABC. Find the image of the vertices
of ∆ABC under the combined rotation of - 180° followed by +90° about the centre
origin. Draw ∆ABC and its image in the same graph paper. Write a single transformation
of above combined transformation.
7. Let M(1, 1), N(-3, -6) and P(6, -1) be the vertices of ∆MNP. The vertices are transformed
by a single transformation obtained by the combination of the rotation [(0,0), + 90°]
and on the same direction of a rotation [(0,0), + 180°]. Find the coordinates of the
images of these points.
8. Draw a triangle with vertices A(1, 2), B(-2, 3) and C(2, 4) on a graph paper.
(a) Find the image ∆A'B'C' under rotation of - 90° about origin.
(b) Find the image ∆A"B"C" of ∆A'B'C' under rotation through+ 180° about the point (2.2).
(c) Find the single transformation to represent these two rotations.
9. If R1 represents a rotation of +90° about the origin and R2 represents a rotation of
- 270° about the origin. What transformation does R2oR1 represents ? Transform the
triangle with vertices A(-4, 0), B (-6, 2) and C(-4, 2) under R2oR1. Draw graphs of ∆ABC
and its images on the same graph paper.
( )10. ∆ABC having vertices A(3, 4), B(-2, 1) and C(4, 1) is translated by T = -1 . The
2
image ∆A'B'C' so formed again rotated through +90° about origin. Find the vertices of
final image. Draw graph of ∆ABC, ∆A'B'C' and ∆A"B"C" on the same graph paper.
11. ∆PQR having vertices P(-2, 2), Q(2, 2), and R(6, 6) is rotated through +90° about the
( )origin. The image ∆P'Q'R' so formed is again translated by T = 3 . Find the vertices
2
of ∆P'Q'R' and ∆P"Q"R". Draw graphs of ∆PQR and its images ∆P'Q'R' and ∆P"Q"R" on the
same graph paper.
Project Work
12. Search some examples of combination of a rotation followed by another rotation.
Prepare a report about them and present in your class room.
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1. (a) (i) (3, -2) (ii) (3, -2) (b) R[(0, 0), 270°]
2. (a) (i) (-4, -5) (ii) 3, 2) (iii) (-4, -6) (b) (i) R[(0, 0), 180°]
(ii) R[(0, 0), 180°] 3.(a) (i) (-5, 4) (ii) (5, 0) (b) (8, -1)
(ii) (-1, -6) (c) (-1, -3)
4. (a) (i) (-5, 5) (ii) A'(-8, 6), B'(5,1) (b) A'(-4, 0), B'(-8, -4)
5. A'(-5, 2), B'(-5, 6), C'(3, 2) A"(5, -2), B"(5, -6), C"(-3, -2)
6. A'(2, -5), B'(5, -4), C'(4, -8), R[(0, 0), -90°] 7. M'(1,-1), N'(-6,3), P'(-1,-6), R[(0, 0), 270°]
8. (a) A'(2, -1), B'(3, 2), C'(4, -2) (b) A"(2, 5), B"(1, 2), C"(0, 6)
(c) R[(0, 4), +90°] 9. R[(0, 0), -180], A'(4, 0), B'(6, -2), C'(4, 2)
10. A'(2, 6), B'(-3, 3), C'(3, 3), A"(-6, 2), B"(-3, -3), C"(-3, 3)
11. P'(-2, -2), Q'(-2, 2) and (-6, 6), P"(1, 0), Q"(1, 4), R"(-3, 8)
11.4 Combinations of Reflection and Reflection
A point or an object can be reflected and the image so formed can further be reflected. There
are two cases in combination of reflections.
A. When the axes of reflections are parallel Y
Let x = - 2 and x = 2 be two parallel lines represented l1 l2
by l1 and l2 respectively.
Let us take a point P(-5, 2). Firstly it is reflected on l1 i.e. (–5, 2) M P' N P"
x = -2 line we get, P O
P (–5, 2) x = –2 P'(1, 2) X' X
Again P'(1, 2) is reflected on l2 i.e. x = 2 line x = –2 x=2
P' (–5, 2) x = 2 P" (3, 2)
Hence, P (–5, 2) P" (3, 2) Y'
o
Let us take two points M on l1 and N on l2 and find MN .
o 4
0
( ) ( )Here, M(-2, 2) and N(2, 2) MN =
2+2 = .
2-2
o 4 = 8 .
0 0
( ) ( )But total displacement is 2 MN = 2
Also, P (–5, 2) T= 8 P' (–5 + 8, 2 + 0) = PI (3, 2)
0
But P"= P'
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This shows that combination of two reflections when the axes of reflections are parallel
is equivalent to a translation and the translation is twice the distance between the axes of
reflection and the direction is perpendicular to the axes of reflections.
Let l1 and l2 be two parallel lines which are represented by x = x1 and x = x2 respectively.
P is reflected on l1 line, we get,
P' x = x1 P"
Again p'is reflected on x = x2 and we get,
P' x = x2 P"
By definition of reflection,
PM = MP' , P'N = NP"
PP' = 2MP' and P'P" = 2P'N
o oo
Now, PP" = PP' + P'P"
oo
= 2 MP' + 2 P'N
oo
= 2( MP' + P'N )
( )= 2 x2-x1 , ( y2 - y1 = 0)
0
Hence if the axes of reflections are parallel, a reflection followed by another reflection is
equivalent to the translation. The translation is twice the distance between the axes of
reflection and the direction is perpendicular to the axes of reflection.
B. When the axes of reflections intersect at a poin. Y y=x
P (2, 6)
Let us consider two axes of reflection x-axis and A
line y = x. The angle between them is 45°.
Let us take P(2, 6) in the plane. Firstly P(2, 6) is
reflected on y = x line.
P (2, 6) y=x P' (6, 2)
Again P'(6, 2) is reflected on x - axis. P' (6, 2)
BX
P' (6, 2) x-axis P" (6, –2) X' T P"(6, –2)
O
Now, measure POP" we get POP" = 90° (-ve).
Y'
The point of intersection of y = x and x - axis is the
origin.
Hence, the combined transformation for P(2, 6) is
P' (2, 6) P' (6, –2)
? Combined transformation is equivalent to rotation and the angle of rotation is twice the
angle between the axes. The centre of rotation is the point of intersection of the axes.
Let l1 and l2 be two lines intersecting at O. Let the lines be the axes of reflections. The angle
between them is T i.e. AOB = T
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Let P be reflected on line l1, we get,
P P'
Again P' is reflected on l2 we get,
P' P"
Now, join OP, OP' OP"
By simple plane geometry POA = AOP' , P'OB = BOP"
POP" = POP' + P'OP"
= 2AOP' + 2P'OB
= 2(AOP' + P'OB) = 2T
Hence, if the axes of reflections intersect at a point, then combined reflection is equivalent to
a rotation about the centre, the point of intersection of axes, the angle of rotation is the angle
between the axes. The direction of rotation is direction from first axis (OA) to the second
axes (OB).
Worked out Examples
Example 1. Let r1 denote the reflection on line x = 2 and r2 denote the reflection on
Solution: line x = 4. Then find the image of the point P(1, 5) under the combined
transformation r2or1 and r1or2.
Here, r1 : x = 2 and r2 : x = 4 are parallel axes of reflection, So, the combined
reflection is equivalent to translation. For r2 o r1 .
( ) ( )translation vector T = 2 4-2 = 4
0 0
? P(1, 5) T = 40
P' (1 + 4, 5 + 0) = P' (5, 5)
( ) ( )For r2 o r1, translation vector T is given by T = 2 2-4 = -4
0 0
Image of P is given by,
P(1, 5) T = -04 P' (1 – 4, 5 + 0) = P' (–3, 5)
Alternatively
For r2 o r1 , first reflect P on r1 i.e. x = 2 Y
P P' P"
P(1, 5) x=2 P' (2 × 2 - 1, 5) = P'(3, 5)
and then P'(3, 5) is reflected on r2 i.e. x = 4
P' (3, 5) x=4 P"(2 × 4 - 3, 5) = P"(5, 5) X' O r1 r2 X
x=4
For r1o r2, first reflect P on r2 i.e. x = 4
x=2
P(1, 5) x = 4 P'(2 × 4 - 1, 5) = P'(7, 5)
Y'
Again P'(7, 5) is reflected on r1 i.e. x = 2
P'(7, 5) x=2 P'(2 × 2 - 7, 5) = P'(-3, 5)
Note : r1 o r2 z r2 o r1
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Example 2. If r1 represents the reflection on x-axis and r2 represents the reflection on y
Solution: axis, find the image of the point P(2, 5) under the combined transformation
r2 o r1
Example 3.
Solution: Here, r1 = reflection on x-axis
r2 = reflection on y-axis
X - axis and Y - axis intersect at the origin and the angle between them is 90°.
Hence, the combined reflection is rotation of (2 × 90) = 180° about O.
P(2, 5) R[(0, 0),180°] P'(-2, -5).
Alternatively
r2o r1 means first reflection on r1 (x - axis), then reflection on r2 (y - axis)
P (2, 5) x-axis P'(2, -5)
P' (2, - 5) y-axis P" (-2, -5)
A triangle with vertices A(1, 2), B(4, -1) and C(2, 5) is reflected successively
on the the lines x = 5 and y = -2. Find by stating the coordinates and
graphically represent the images under these transformations. State also
the single transformation given by the combination of these transformation.
Here, A(1, 2), B(4, -1) and C are the vertices of ∆ABC.
Reflecting the ∆ABC on line x = 5, we get,
P(x, y) x=h P'(2h – x, y) x=5
Y
A(1, 2) x=5 A'(2 × 5 - 1, 2)
C P" C'
= A'(9, 2)
B(4, -1) B'(2 × 5 - 4, -1) A A'
X
= B'(6, -1) X'
O B B'
C(2, 5) C'(2 × 5 - 2, 5) P y = –2
B"
= C'(8, 5)
Y' A"
Again reflecting ∆A'B'C' on y = -2, we get.
C"
A'(9, 2) y = –2 A"(9. 2 × (-2) - 2)
= A" (9. -6)
B'(6, -1) B"(6, 2 ×(-2) + 1)
= B"(6, - 3)
C'(8, 5) C"(8, 2 × (-2)- 5)
= C"(8, - 9)
∆ABC, ∆A'B'C' and ∆A"B"C" are plotted in the graph. The lines x = 5 and
y = -2 intersect at point P(5, -2). The angle between these two lines is 180°
Hence, the combined reflection is rotation of 180° about centre P(5, -2)
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Example 4. A triangle with vertices P(3, 5), Q(4, 1) and R(1, 0) is reflected successively
on the lines x =2 and x =6. Find by stating the coordinates and graphically
represent the images under transformations. Also write a single
transformation representing the above combined transformation.
Solution: Here, P(3, 5), Q(4, 1) and R(1, 0) Y x=6
are the vertices of ∆PQR. x=2
P"
Reflecting the ∆PQR on line x = 2, P' P
we get, P(x, y)x = h P'(2h - x, y)
P(3, 5) x = 2 P"(2 × 2 - 3, 5) X' Q' R Q"
O Q R' R" X
= P'(1, 5)
Q(4, 1) x = 2 Q'(2 × 2 - 4, 1)
= Q'(0, 1) Y'
x = 2 R'(2 × 2 - 1, 0)
R(1, 0)
=R'(3, 0)
Again, reflecting ∆P'Q'R' on the line x = 6, we get
P'(1, 5) x = 6 P"(2 × 6 - 1, 5) = P"(11, 5)
Q(0, 1) x = 6 Q"(2 × 6 - 0, 1) = Q" (12, 1)
R'(3, 0) x = 6 R"(2 × 6 - 3, 0) = R"(9, 0)
Since the axes of reflections are parallel, the combined reflection is a
translation, the translation is two times the displacement between the
( ) ( )axes. Here, translation T is given by 6-2 8
T1 = 2 0 = 0 .
Combined Transformation of Reflection and Translation
( )Let P(x, y) be any point on the plane, T = a be a translation vector. Then,
b
a
P(x, y) T= b P'(x + a, y +b)
Again P' is reflected on x-axis, we get,
P'(x + a, y + b) x-axis P" (x + a, - y - b)
This transformation is translation followed by reflection. Similar, transformation result can
be derived for reflection followed by translation.
i.e. P(x, y) x-axis P'(x, - y) T = ba P"(x + a, -y + b).
Composition of Reflection and Rotation
A point or an object once reflected can further be rotated to get a new image. This is also a
type of combined transformation.
Let P(x, y) be any point on the plane. Then P is reflected on x - axis, we get,
P(x, y) P'(x, - y)
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Again P' is rotated through -90° about origin.
P'(x, -y) P"(-y, -x)
Similar, combination of transformation results can be derived for rotation followed by
reflection.
Let, Rx = reflection on x-axis
Ry = reflection on y - axis
Ry = x = reflection on line y = x
Ry = -x = reflection on line y = - x
R1 = rotation through 90° about origin
R2 = rotation through - 90° about origin
R3 = Rotation through 180° about origin.
Then, discuss about the following combination of transformation.
(i) Rx o R1 (ii) Ry o R2 (iii) Ry ox R1 etc.
Example 5. Let Rx denote reflection on x - axis and R1 denote, the rotation of +90° about
Solution: the origin. Find the image of ∆PQR with P(2, 3), Q(6, 7) and R(0, 3) under
R1oRx .
P(2, 3). Q(6, 7) and R(0, 3) are the vertices of ∆PQR. R1 o Rx means Rx followed
by R1, where
R1 = rotation of +90° about O.
Rx = reflection on x-axis.
Under reflection on x-axis, we get,
P(x, y) x-axis P'(x, –y)
P(2, 3) P'(2, -3)
Q(6, 7) Q'(6, -7)
R'(0, 3) R"(0, -3)
Again, ' P'Q'R' is rotated through +90° about origin.
P(x, y) R[(0, 0), +90°] P'(–y, x)
P'(2, –3) P"(3, 2)
Q'(6, -7) P"(7, 6)
R'(0, -3) P"(3, 0)
Exercise 11.3
Short Question
1. (a) Write the combined reflection of reflection in x = 2 followed by reflection in x = 6
(b) Write the combined reflection of reflection in x = 0 followed by reflection in y = 0
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(c) Write the combined reflection of reflection in x = 4 followed by reflection in y = -2
(d) Write the combined reflection for reflection an y = x followed by reflection x - axis.
(e) What is the combined reflection when the axes of reflection are (i) parallel (ii)
intersecting at a point?
2. Let Rx = Reflection in x - axis
Ry = Reflection in y - axis
R1 = Reflection in line y = x
R2 = Reflection on y = -x
( )T =2 , translation vector.
3
r1 = rotation of +90° about origin.
r2 = rotation of -90° about origin
r3 = rotation of 180° about origin.
Find the image of points under the following combined transformations:
(a) RxoRy (2, 3) (b) RyoRx(2, 3) (c) R1oR2(2, 4)
(d) R2oR1(6, 2) (e) R1oT(2, 4) (f) ToRx(4, -4)
(g) r1or2(4, 5) (h) r2or1(-4, 5) (i) r1oRx(6, 8)
(j) T1oR2(2, 4) (k) T1o r2(-4, -5) (l) ToRx(-5, -5)
Long Questions
3. A(- 4, 0), B(-6, 2) and C(-4, 3) are the vertices of ∆ABC. The triangle ABC is reflected
successively on the line x = - 3 and x = 1. Find the final image and describe the single
transformation equivalent to the combination of these transformation.
4. If R is the reflection on the line y = x and Ry is the reflection the line x = 0. Then what
does RoRy represent ? By using the combined transformation of this result, find the
image of ∆ ABC whose vertices are A(3, 4), B(7, 0), and C(4, 0). Draw graph of ∆ABC
and its image on the same graph paper.
5. On a graph paper draw ∆ABC having the vertices A(5, 4), B(2, 2) and C(5, 2). Find the
image of ∆ABC by stating coordinates and graphing them after successive reflections on
x-axis following by reflection on the line y = x.
6. A triangle with vertices P(1, 4), Q(4, 1) and R(7, 5) is reflected successively in
y- axis followed by reflection on the line y = x. Find the final image of ∆PQR and
present the object and images in the same graph paper.
7. A triangle with the vertices A(1, 2), B(4, -1) and C(2, 5) is reflected successively in the
line x = 4 and y = -2. Find by stating coordinates and graphically represent the images
under there transformations given by the combination of these transformation.
8. A(4, 5), B(-2, -2) and C(4, 8) are the vertices of ∆ ABC. ∆ABC is rotated through +90°
about the origin followed by the reflection in x-axis. State the single transformation.
Draw ∆ABC and its images in the same graph paper.
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9. If R1 denotes rotation of +90° about origin and R2 denotes rotation of -270° about the
origin. What does R2oR1 represent? By using R2oR1, transform quadrilateral ABCD with
A(-4, 0), B(-6, 2), C(-2, 4) and D(-1, 1). Present the quadrilateral ABCD and its image on
the same graph paper.
10. O(0. 0) A(2, 0), B(3, 2) and C(1, 2) are the vertices of quadrilateral OABC. Translate the
0
quadrilateral by translation vector 2 .
Reflect the image so formed on the line x = 3. Represent the images and object in the
same graph.
11. A triangle having vertices A(2, 5), B(-1, 5) and C(4, 1) is rotated through +90°
about the origin. The image obtained is reflected on the line x = 0. Find the vertices
of image triangles. Show all the triangles in the same paper and also write the single
transformation to represent these two transformations.
12. State the single transformation equivalent to the combination of reflections on the
x-axis and y-axis respectively. Using this single transformation find the coordinates
of the vertices of the image of ∆ABC having vertices A(2, 3), B(3, -4) and C(-1, 2). Also
draw the object and its image on the same graph.
13. Draw ∆ABC having the vertices A(2, 0), B(3, 1) and C(1, 1) on a graph paper. It is rotated
about the origin through -90° and present ∆A'B'C' on the same graph paper. Then '
A'B'C' is reflected on the line x – y = 0 and plot 'A"B"C" on the same graph paper. Write
down the coordinates of vertices of ∆A"B"C".
Project Work
14. Search some examples of combination of reflection and their use in daily life. Prepare
a report and present in your classroom.
1. (a) T = 8 (b) R[(0, 0), 180°] (c) R[(4, -2), 180°]
0 (e) (i) Translation (ii) Rotation
(d) R[(0, 0), -90°]
2. (a) (-2, -3) (b) (-2, -3) (c) (-2, -4) (d) (-6, -2)
(e) (7, 4) (f) (6, 7) (g) (4, 5) (h) (-4, 5)
(i) (-8, 6) (j) (-2, 1) (k) (-3, 7) (l) (-3, 8)
3. 8 , A'(4, 0), B'(2, 2), C'(4, 3) 4. R[(0, 0), -90°], A'(4, -3), B'(0, -7), C'(0, -4)
0
5. R[(0, 0), +90°, A'(5, -4), B'(2, -2), C'(5, -2), A"(-4, 5), B"(-2, 2), C"(-2, 5)
6. P'(-1, 4), Q'(-4, 1), R'(-7, 5), P"(4, -1), Q"(1, -4), R"(5, -7)
7. A'(7, 2), B'(4, -1), C'(4, 8), A"(7, -6), B"(4, -3), C"(4, -12)
8. A'(-5, 4), B'(2, -2), C'(-8, 4), A"(-5, -4), B"(2, 2), C"(-8, -4), y = x
9. A'(0, -4), B'(-2, -6), C'(-4, -2), A"(4, 0), B"(6, -2), C"(2, -4), R[(0, 0), -180°]
10. O'(6, 2), A'(2, 2), B'(3, 4), C'(1, 4), O"(6, 2), A"(4, 2), B"(3, 4), C"(5, 4)
11. A'(-5, 2), B'(5, - 1), C'(-1, 4), A"(5, 2), B"(5, -1), C"(1, 4)
12. R[(0, 0), 180°], A'(-2, -3), B'(-3, 4), C'(1, -2)
13. A'(0, -2), B'(1, -3), C'(1, -1), A"(-2, 0), B"(-3, 1), C"(-1, 1)
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11.5 Combination of Enlargement and Enlargement
An object once enlarged (or reduced) can again be enlarged. The single enlargement (or
reduction) is equivalent to these two transformations is the combination of enlargements
and enlargement. There are two cases in the combination of two enlargements (or reduction)
A. When the centres of enlargements are same Y A"
Let AB be a given line segment and E1[O, k1] be an A'
enlargement with centre O and scale factor k1. Let A'B' be
the image of a given line segment AB under enlargement
E1. Then we write A'B' = k1AB A B"
Again let E2 [O, k2] be another enlargement with the B B'
same centre O and scale factor K2. Let A"B" be the image
of A'B' under enlargement E2. Then we write X' O X
A"B" = k2A'B' Y'
Finally, we get, A"B" = k1 A'B' = k1. k2 AB
Here, A"B" is the image of AB under the enlargement E[O, k1 k2].
Hence, we conclude that an enlargement with centre O and scale factor k1, followed
by another enlargement with the same centre O and scale factor k2 is equivalent to the
enlargement with the same centre with scale factor k1.k2 .
Note :
(i) If E1[(0, 0), k1] and E2[(0, 0), k2] are two enlargements, the combined enlargement is
E[(0, 0), k1k2].
(ii) If E2[(a, b), k1] and E2[(a, b), k2] are two enlargements, the combined enlargement is
E[(a, b), k1k2].
Example 1. If P(3, 2), Q(9, 2) R(9, 6) and S(3, 6) are the vertices of square PQRS. Find the
Solution: image of square PQRS under an enlargement with the centre at origin and
scale factor 2 followed by an enlargement with centre at origin and scale
factor 3.
The enlargement E1[0, 2] followed by enlargement E2 [0, 3] is equivalent to
the single enlargement E[0, 2 × 3] = E[0, 6] with the same centre O and scale
factor 6.
Then, the image of square PQRS due to E[0, 6] is P'Q'R'S'
Object Image
P(x, y) P'(kx, ky)
P(3, 2) P'(6 × 3, 6 × 2) = P'(18, 12)
Q(9, 2) Q'(6 × 9, 6 × 2) = R'(54, 12)
R(9, 6) R' (6 × 9. 6 × 6) = R'(54, 36)
S(3, 6) S'(6 × 3, 6 × 6) = S'(18, 36)
squares PQRS squares P'Q'R'S'.
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B. When the centres of enlargements are different. A" B"
Let us take a line segment AB and E1[O, k1] be the enlargement
with centre P and scale factor k1. A' B'
Then, A'B' = kAB
AB
Again, let E2[O', k2], then A"B" = k2A'B' = k2 k1AB. O O" O'
Join AA" and BB"
Fig. Combined enlargement with
Let the point of intersection of AA" and BB" be O". different centres.
Hence, A"B" is the image of AB under the enlargement
E[O", k1k2] with the third centre O" and scale factor k1k2 .
Example 2. Find the image of ∆ABC with vertices A(1, 1), B(2, 4) and C(2, -1) under
Solution: enlargement E1[(2, 3), 2] followed by another enlargement E2 [(3, 3), 3]
Under enlargement E1 [(2, 3), 2]
We have, P(x, y) E[(a, b), k] P'(kx - ka + a, ky - kb + b)
Now, A(1, 1) E1[(2, 3), 2] A'(2 × 1 - 2 × 2 + 2, 2 × 1 - 2 × 3 + 3) = A'(0, -1)
B(2, 4) B'(2 × 2 - 2 × 2 + 2, 2 × 4 - 2. 3 + 3) = B'(2, 5)
C(2, -1) C'(2 × 2 - 2 × 2 + 2, 2 × (-1) - 2 × 3 + 3) = C'(2, -5)
Again under E2 [(3, 3), 3], ∆A'B'C' is enlargement.
A'(0, -1) A"(3 × 0 - 3 × 3 + 3, 2 × -1 - 3 × 3 + 3) = A' (-6, -8)
B'(2, 5) B"(3 × 2 - 3.3 + 3, 3 × 5 - 3 × 3 + 3) = B"(0, 9)
C'(2, -5) C"(3 × 2 - 3 × 3 + 3, 3 × (-5) - 3 × 3 + 3) = C" (0, -21)
Combination of a transformation with enlargement :
Here, Combination of transformations may be
- Reflection and Enlargement
- Rotation and Enlargement
- Translation and Enlargement and other.
Combination of Reflection and Enlargement
Let P(x, y) be a point an the plane.
Let P be reflected on y-axis. P'(-x, y)
P(x, y) y-axis
Again, let P'(-x, y) be enlarged under E[(0, 0), k]. Then we write
P'(-x, y) P"(-kx, ky)
So, P"(-kx, ky) is the image of P(x, y) under reflection on y-axis followed by enlargement
E[(0,0), k]
Combination other reflections and enlargement can be derived in the similar ways.
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Combination of Rotation and Enlargement
Take a point P(x, y) on the plane.
Let P(x, y) be rotated through +90° about origin.
P(x, y) P'(-y, x)
Again let P' (-y, x) be enlarged under E [(0,0), k], we get,
P'(-y, x) P" (-ky, kx)
P" (-ky, kx) is the image of P(x, y) under rotation +90° about O combination of other rotations
followed by E[(0, 0), 3] with enlargement can be derived in similar ways.
Combination of Translation and Enlargement
Let P(x, y) be a point on the plane.
( )T =a be a translation vector then image of P under T is given by
b
P(x, y) P'(x + a, y + b)
Again P' is enlargement under E[(0, 0), k], we get,
P'(x + a, y +b) P" [k(x + a), k(y + b)]
P"[k (x + a), k(y + b)] is the image of P(x, y) under translation followed by enlargement E[(0, 0), k]
Example 3. A point P(7, 8) is reflected on the line y = -x. The image so obtained is
Solution: enlarged under E[(0,0), 2]. Find the coordinates of the images.
Given point is P(7, 8)
Under the reflection about the line y = - x, we get
P(x, y) y = -x P'(-y, -x)
P(7, 8) P'(-8, - 7)
Again P(-8, -7) is enlarged under E[(0,0), 2],
P'(-8, -7) P"[2 (-8), 2(-7)] = P"(-16, -14)
Example 4. A point P(-2, - 3) is rotated through +90° about origin. The image so obtained
Solution: is enlarged under E[(o,o), 2].
Example 5. Given point is P(-2, - 3),
Solution:
Under rotation through +90° about origin, we get,
P(x, y) P'(-y, x)
P(-2, -3) P'(3, -2)
Again, P'(3, 2) is enlarged under enlargement E[(o, o), 2]
P'(3, -2) P"(6, - 4)
A(1, 0), B(0, 2), and C(2, 1) are the vertices of ∆ABC. Enlarge ∆ABC under
E[(o,o), 3] and the image so formed is again reflected on y-axis. Draw the
graph of ∆ABC and its images on the same graph paper.
Here, A(1, 0), B(0, 2), and C(2, 1) are the vertices of ∆ABC. ∆ABC is enlarged
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under E[(0,0), 3] we get,
P(x, y) E[(0, 0), k] P'(kx, ky)
A(1, 0), o A'(3 × 1, 3 × 0) = A'(3, 0)
B(0, 2) o B'(0, 3 × 2) = B'(0, 6)
C(2, 1) o C'(3 × 2, 3 × 1) = C'(6, 3)
Again, ∆A'B'C' is reflected on Y - axis.
A'(3, 0) o A"(-3, 0)
B'(0, 6) o B"(0, 6)
C'(6, 3) o C"(-6, 3)
Y
B" B'
C" B C'
X' C
X
A" O A A'
Y'
Exercise 11.4
Very Short Questions
1. (a) Transform point A(4, 5) under enlargement E[(0, 0), 2].
(b) Transform point G(2, 2) under enlargement E[(0, 0), -2].
(c) Find the image of P(5, 3) under enlargement E[(1, 1), 2].
(d) Find the image of M(2, -2) under enlargement E[(0, 4), - 2].
2. (a) Let E1[(0,0)2] and E2 [(0,0), 3] what is the single enlargement due to E1oE2?
(b) Let E1 [(0,0), 4] and E2 (0, 0), 1 , what is the single enlargement due to E2E2?
2
(c) Let E1 be [(2, 1), 3], and E2 be [(2, 1), 2]. Write the single transformation equivalent
to E2E1.
Short Questions
3. (a) Let E1[(0,0), 2] and E2 [(0, 0), 3] be two enlargements. Find the image of P(2, 5)
under E2oE1 .
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(b) Let E be [(1, 1), 3] and E2 be [(1, 1), 2]. Find the image of P(2, 6) under E2oE1.
(c) Find the image of N(2, 3) under enlargement E1[O, 2] followed by enlargement
1
E2 O, 2 2 .
(d) A point P(3, k) is the first transformed by E1[O, 2] and by E2 O, 3 so that the final
image is [9, 12], find the value of k. 2
4. Let E1 = enlargement about (0, 0) with scale factor 2.
E2 = enlargement about (0, 0) with scale factor 3 .
2
R1 = Rotation of + 90° about 0.
R2 = Rotation of 180° about 0.
[ ]T = 2 , translation vector.
3
Find the image of the following points the combined transformation given below:
(a) E1 o E2 (2, 4) (b) E1 oT (4, 5) (c) E2 o T(2, 5)
(d) ToE1 (-2, -3) (e) R1 o E1 (4, 5) (f) R2 oE2 (1, 2)
(g) E2 o R1 (4, 4)
Long Questions
5. P(3, 0) , Q(4, 2), R(2, 4), and S(1, 2) are the vertices of parallelogram PQRS. Draw PQRSon
graph paper. Enlarge the parallelogram under E1 [(0,0), 2] followed by E2 (0, 0), 1 1 .
2
Find the images P'Q'R'S' and P"Q"R"S". Draw P'Q"R'S' and P"Q"R"S" on the same graph paper.
6. The vertices of ∆ABC are A(2, 3), B(4, 5), and C(6, 2). If E1 [(0, 0),2] and E2 [(0, 0)2] are
two enlargements, find the coordinates of the image of ∆ABC under the enlargement
E2oE1. Draw both triangles on the same graph.
7. Find the image of ∆ABC with the vertices A(2, 1), B(3, 5) and C(5, 4) under enlargement
E1 [(0, 0), 3/2] followed by another enlargement E2 [(0,0), -2]. Draw ∆ABC and its images
on the same graph paper.
8. P(3, 0), Q(4, 2), R(2, 4) and S(1, 2) are the vertices of parallelogram PQRS. Enlarge the
parallelogram PQRS E1 [(1, 2), 2] and by enlargement E2 [(1, 2), 3]. Draw both the object
triangle and image triangle on the same graph paper.
9. A(2, 5), B(-1, 3), and C(4, 1) are the vertices of ∆ABC. Find the coordinates of the
image of the ∆ABC under the rotation of positive 90° about the origin followed by the
enlargement E[(0,0), 2]
10. Let E denote enlargement about centre (-3, - 4) with scale factor 2 and R denote a
reflection of the line y = x. ∆ABC with vertices A(2, 0), B(3, 1), and C(1, 1) is mapped
under the combined transformation RoE. Find the image of ∆ABC and draw both figures
on the same graph.
11. A(2, 0), B(3, 1), and C(1, 1) are the vertices of a triangle ABC. Find the coordinates of
the vertices of the triangle ABC under the reflection on the line x = y following the
enlargement E[(-3, -4), 2]
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-2
( )12. ∆PQR having the vertices P(3, 4), Q(2, 1), and R(4, 2) is translated by T = 3 . The
image so formed is enlarged by E[(0, 0), 2]. Writing the coordinates of the vertices of
images thus obtained of the ∆PQR and its images in the same graph paper.
13. The vertices of ∆ABC are A(1, 1), B(3, 1), and C(2, 3) ∆A'B'C' is obtained by rotating
∆ABC about the origin half turn. The image formed is enlarged about the same centre
with scale factor 2 to get ∆A"B"C". Find the coordinates of the final image and represent
all three triangles on the same graph paper.
14. M(3, 4), N(1, 1), and P(4, 1) are the vertices of ∆MNP. Find the image of ∆MNP. Under
the enlargement with centre (1, -1) and scale factor -2 followed by the rotation about the
origin through negative quarter turn. Also show the images on the same graph paper.
1. (a) A'(8, 10) (b) G'(-4, -4) (c) P'(9, 5) (d) M'(-4, 16)
2. (a) E[(0, 0), 6] (b) E[(0, 0), 2] (c) E[(0, 0), 6]
3. (a) P'(12, 30) (b) P'(7, 31) (e) N'(10, 15) (d) k = 4
4. (a) (6, 12) (b) (12, 16) (c) (8, 16) (d) (-2, -3)
(e) (-10, 8) (f) - 3 , -3 (g) (-6, 6)
2
5. P'(6, 0), Q'(8, 4), R'(4, 8), S'(2, 4), P"(9, 0), R"(6, 12), Q"(12, 6), S"(3, 6)
6. A'(8, 12), B'(16, 20), C'(24, 8) 7. A"(-6, -3), B"(-9, -15), C'(-15, -12)
8. P"(13, -10), Q"(19, 2), R"(7, 14), S"(1, 2)
9. A'(-5, 2), B'(-3, -1), C'(-1, 4), A"(-10, 4), B"(-6, -2), C"(-2, 8)
10. A'(7, 4), B'(9, 6), C'(5, 6), A"(4, 7), B"(6, 9), C"(6, 5)
11. A'(0. 2) . B'(1, 3), C'(1, 1), A"(3, 8), B"(5, 10), C"(5, 6)
12. P'(1, 7), Q'(0, 4), R'(2, 5), P"(2, 14), Q"(0, 8), R"(4, 10)
13. A'(-1, -1), B'(-3, -1), C'(-2, -3), A"(-2, -2), B"(-6, -2), C"(-4, -6)
14. M'(-3, -11), N'(1, -5), P"(-5, -5), M"(-11, 3), N"(-5, -1), P"(-5, 5)
11.6 Inversion Transformation and Inversion Circle
Introduction
Let us discuss answer of the following questions in group.
(a) What is the equation of a circle with centre at the origin and radius 5 units?
(b) Find the centre and radius of the circle whose equation is C
OB
(x - 4)2 + (y + 3)2 = 25 ? A
(c) In the given circle O is the centre of a circle, find the value of ACB.
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(d) In the given circle, O is the centre and P is the point of contact. What is A
the relation between OP and AB ?
OP
(e) What are the meanings of similarity and line of symmetry ? B
11.7 Inversion Transformation
Transformation is a type of function. So, it may also have its inverse. The inverse of a
transformation is called its inverse. If T is a transformation on a set A to itself which carries
each element 'a' of A into a corresponding element 'b' of A. Then, the transformation that
reverses transformation T, by carrying each element 'b' of A back into its original element 'a'
of A is called inverse transformation. Then, it is denoted by T–1. We can write:
T:aob
T–1 : b o a
The inverse of reflection is the reflection on the same line of reflection. Similarly, the inverse
of translation is a transformation in the same magnitude but opposite in direction. The size
1
of transformation of magnitude p and p , having the same centre are inverse to each other.
Inversion transformation does not preserve collinearity and distance. It can be generalized
as reflection in a line.
Concept of Inversion Circle and Inversion Point
O P=P' OP P' O P' P
Fig (i) Fig (ii) Fig (iii)
In all figure O is the centre of circle, r is radius. In figure (i), the points P and P' are on the
circumference of the circle.
In figure (ii) P lies inside the circle and P' lies outside the circle.
In figure (iii) P' lies inside the circle and P lies outside the circle.
In each of above cases OP. OP' = r2 is true. We can observe that P o P', P' o P the points P
and P' are called relative inverse of each other.
Inversion in a circle is a method of transformation of A
geometric figure into other figure. It is like process of D
reflection on a line. Since the reflection on a line depends O P' P'
on the particular line chosen, inversion in a circle depends
on the particular circle. Let D be a circle with centre O and
radius r and P are any point other than origin. Then, P' is B
called inversion of P.
Inversion is the process of transformation of point P to corresponding set of points P' called
in the inverse points.
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Let D be a fixed circle. Then, in the above figure, O is the inversion centre, D is called
inversion circle, AB is called polar and P is the pole.
For every point P other than O, we define a point P' is the inverse point of P. It is to be noted
that if P is inside the circle P' is outside of the circle and vice-versa.
Features of Inversion Circle
The following are the some important features of inversion circle.
(a) To each point P of the plane other than O, there corresponds an inverse point P'.
(b) If P lies on the circumference of the inversion D, then P = P'
(c) If OP < r, i.e. if P lies inside there circle P' lies outside the circle EFG. Then
OP' > r. Conversely if OP > r, i.e. if P lies outside the circle, P' and inverse point
P' lies inside the circle EFG.
(d) The point P and the inverse point P' can always be interchanged.
(e) If P' is the image of P, P is the image of the P'. Hence, the point and its inverse point
can be always be interchanged. This property can be referred as symmetric feature
of inversion. i.e., (P')' = P
11.8 Construction of Inverse Point Geometrically
There may arise three cases.
(a) When the point P lies on the circumference of the circle then the point O
P' is the inverse point itself. P' = P
(b) When the point P is inside the circle D. Draw D U P'
o OP
a ray OP passing through O and P. A chord UV V
is drawn perpendicular to the ray OoP passing
through the point P. Then tagents UP' and VP'
are drawn to the circle at the points U and V. OU
and OV are joined. The point of intersection of
the tangents UP' and VP' is P' which lies on the
ray OoP. Join OU and OV. Then OPU and OUP'
will be right angles. In right angled triangles
∆OPU and ∆OP'U, we have,
OPU = OUP' (Both are right angles)
UOP = UOP' (Common angle)
OUP = UP'O (the remaining angles)
Hence 'OPU and 'OP'U are similar.
Therefore, from similar triangles ∆OPU and ∆OP'U. we have,
OP = OU or. OP.OP' = OU2
OU OP'
? OP.OP' = r2(i.e. OU = r)
Hence, the point P' is the inverse of point P.
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P
(c) Let the P be outside the inversion circle O P'
D. Join OP. Let M be the mid point of V
OP. Draw another circle E about centre
M and radius OM = MP. Then the two
circles D and E intersect at the points U
and V. Join UP and VP. Then OUP and
OVP are inscribed in semi-circle E. D
Hence, they are right angles. Again UP
and VP are tangents to the circle D. The
chord UV meets OP at P'. The point P' is
the inverse of point P. Hence, the point
P' is the inverse of the point P relative to
the circle D.
11.9 Use of Coordinates in Inversion Circle
(a) Centre at the origin (Standard form): P (x, y)
y
The inverse of a point P(x, y) with respect
to a circle with centre at the origin and the
radius r is a point P'(x', y') such that
x' = r2x , y = r2y P'(x',y')
x2 + y2 x2 + y2
Proof: Sx
x' OR
Let O (0, 0) be the centre and r be the radius y'
of inversion circle. Then equation of the
circle is x2 + y2 = r2. Let P'(x', y') be the
inverse point of P(x, y). Then, by definition,
OP.OP' = r2, where,
OP = x2 + y2
OP' = x'2 + y'2
From P and P' draw PSAOX and P'RAOX. Since the points O, P', P are collinear, the
triangles OPS and OP'R are similar. In similar triangles the corresponding sides are
proportional.
OR = P'R = OP'
OS PS OP
x' y' OP' OP
or, xx' = yy' = OP . OP
or, x = y =
OP.OP' = x2 r2 y2
OP2 +
Then,
and x' = r2x
y' = x2 r+2y y2
x2 + y2
Hence the equation of inverse of any point P(x, y) with respect to a circle with centre at
the origin and the radius r is a point P'(x', y) such that
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x' = r2x and y' = r2y
x2 + y2 x2 + y2
Note :
1. When r = 0, x2 + y2 = 0, is the equation of inversion circle. Then x' and y' do not have
any real values, then the center of inverse is a point at infinity.
2. For any point on the circumference of the inversion circle x2 + y2 = r2, its inverse
point P'(x', y') is given by x = x' and y = y'.
(b) Centre at any Point (Central form) y P(x, y)
The equations of inverse of P'(x', y')P'(x', y') S
any point P(x, y) with respect C(h, k) r x'
to a circle with centre at
C(h, k) and radius r is a point R W
P'(x', y') such that O
x' = h + (x – r2 (x – h) k)2 UV
h)2 + (y –
y'
and y' = k + r2 (y – k)
(x – h)2 + (y – k)2
Proof:
x
Let C(h, k) be the centre and r
be the radius of the circle. Then
equation of the inversion circle is
given by
(x – h)2 + (y – k)2 = r2
Let P'(x', y') be the inverse point of P(x, y)
Then by definition, CP.CP' = r2.
Now, CP = (x – h)2 + (y – k)2
CP' = (x' – h)2 + (y' – k)2
Draw CU A OX, P'V A OX, PW A OX. Then also draw CS A PW, CS cuts PV at R. Since
the points C, P' and P are collinear, 'CPS and 'CP'R are similar.
In similar triangles, the corresponding sides are proportional
CR = P'R = CP'
CS PS CP
x' – h y' – k CP'.CP
or, x–h = y–k = CP2
or, x' – h = y' – k = r2
x–h y–k (x – h)2 + (y – k)2
r2 (x – h) r2 (y – k)
Then, x' = h + (x – h)2 + (y – k)2 and y' = k+ (x – h)2 + (y – k)2
Hence, the equation of inverse of any point P(x, y) with respect to a circle with centre
at (h, k) and radius r is a point P'(x', y') such that
x' = h + r2 (x – h) and y' = k + r2 (y – k)
(x – h)2 + (y – k)2 (x – h)2 + (y – k)2
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Worked out Examples
Example 1: From the adjoining figure give the following concepts:
Solutions:
(a) Inversion circle (b) Inversion radius
(c) Centre of inversion (d) Relation between OP, OP' and OA.
From the above figure, N
D
(a) Circle D is the inversion circle A
(b) OA = r, radius of inversion circle
OP P'
R
(c) O is the centre of inversion circle M
(d) By definition of inversion circle
OP.OP' = OA2
or, OP.OP' = r2
Example 2: Find the inverse of the point P, Q, R which are at distances 2, 4, 8 units from
Solutions: the centre of O of the circle with radius 4 units.
Let OP = 2, OQ = 4, OR = 8 units.
Let P', Q', and R' the inverse points of P, Q, and R' respectively.
Now, for inverse point of P.
OP.OP' = r2, r = 4 R
or, 2.OP' = 42
or, OP' = 16 Q
2 Q'
? OP' = 8 units
R'
i.e., P' is at 8 units from O. OP
Similarly, for the inverse point of Q, P'
OQ.OQ' = r2
or, 4 × OQ' = 16
or, OQ'= 4 units
i.e., Q' is at 4 units from O.
Here, OQ' = OQ = r. The point Q itself is its inverse point.
For the inverse point of R.
OR.OR'= r2
or, OR' = 42 = 16 =2
OR 8
? R' is 2 units from O.
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Example 3: Find the inverse of the point (i) A(3, 5) (ii) B(2, – 5) with respect to the circle
Solutions: x2 + y2 = 1.
Example 4: Equation of given circle is x2 + y2 = 1.
Solutions:
radius (r) = 1 unit, centre of circle = O (0, 0)
(i) To find inverse of A(3, 5).
r = 1, x = 3, y = 5. Let A'(x', y') be inverse of A.
x' = r2x = 1×3 = 3 = 3
x2 + y2 32+ 52 9 + 25 34
r2y 12 × 5 5 5
y' = x2 + y2 = 32+ 52 = 9 + 25 = 34
Hence, the required inverse point is A'( 3 , 354 )
34
(ii) To find the inverse of B(2, –5)
r = 1, x = 2, y = 5
Let B'(x', y') be inverse image of B.
Then, x' = r2x
y' x2 + y2
= 12 × 2 = 2
22+ (–5)2 29
= r2y = 12 × (–5) = – 5
x2 + y2 22+ (–5)2 29
Hence, the required inverse of point B is B' 2 , – 5 .
29 29
Find the inverse of the point (i) P(3, 2), (ii) Q(3, 3) with respect to the circle
with centre (2, 3) and radius of inverse circle 2 units.
According question,
radius (r) = 2, centre (h, k) = (2, 3)
(i) To find inverse of point P(3, 2).
r = 2, (h, k) = (2, 3), (x, y) = (3, 2)
Let, P'(x', y') be required inverse of point P.
Then, =h+ r2 (x – h) =2+ 22 (3 – 2)
x' (x – h)2 + (y – k)2 (3 – 2)2 + (2 – 3)2
and y' = 2 + 22 × 1 = 2 + 4 = 2 + 2 = 4
(1)2 + (–1)2 2
=k+ r2 (y – k) =3+ 22 (2 – 3)
(x – h)2 + (y – k)2 (3 – 2)2 + (2 – 3)2
= 3 + 22 × 1 = 3 + –4 = 3 – 2 = 1.
(1)2 + (–1)2 2
Hence, the inverse point of P(3, 2) is P'(4, 1).
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(ii) To find inverse of point Q(3, 3)
r = 2, (h, k) = (2, 3), (x, y) = (3, 3)
x – h = 3 – 2 = 1, y – k = 3 – 3 = 0
Let Q' (x', y') be required inverse of point Q.
x' = h+ r2 (x – h) =2+ 22 × (3 – 2)
(x – h)2 + (y – k)2 (3 – 2)2 + (3 – 3)2
22 × 1
= 2 + (1)2 + (0)2
=2+4
=6
and y' = k+ (x – r2 (y – k) k)2 = 3 + 22 × (3 – 3)
h)2 + (y – (3 – 2)2 + (3 – 3)2
= 3 + 22 × 0 = 3 + 0 = 3.
(1)2 + (0)2
Hence, the inverse point of Q(3, 3) is Q'(6, 3).
Exercise 11.6
Very short questions A P'
1. From the given figure, give the following concepts:
OP
(a) inversion circle D
(b) inversion radius
(c) centre of inversion
(d) relation among OP, OP' and r.
2. Fill in the gaps in the following table:
Distance of the Distance of the
inverse point from
S.N. given point from the Radius of inversion centre of inversion
centre of inversion Circle
circle
circle ..........
..........
(a) 1 2
1
(b) 1 3 2
9
2
(c) 18 .......... 3
625
(d) 16 .......... 1
(e) 3 .......... 4
(f) .......... 12
(g) .......... 25
9
(h) .......... 24
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Short question
3. Find the inverse points of the following points with respect to the inversion circle with
centre at the origin.
S.N. Given points Radius of inversion circle
a P(2, 3) 5
b Q(1, 2) 2
c R(4, 5) 4
d S(3, 0) 3
4. (a) Find the distance of inverse of the points A, B and C which are at distances 3, 6, 9
units respectively from the centre O of the circle with radius 6 units.
(b) Find the inverse of the points P, Q and R which are at distances 2, 4, 8 respectively
units from the centre O of the circle with radius 8 units.
(c) Find the inverse of the points M, N and P which are at distances 2.5, 5 and 10 units
from the centre O of the circle with radius 5 units.
5. (a) Find the inverse of the point (2, –5) with respect to the circle x2 + y2 = 1.
(b) Find the inverse of the point (3, 4) with respect to the circle x2 + y2 = 4.
(c) Find the inverse of the point (4, 5) with respect to the circle x2 + y2 = 25.
6. (a) Find the inverse of the point (6, 2) with respect to a circle centre at the point (1, 2)
of radius 3 units.
(b) Find the inverse of the point (4, 2) with respect to a circle centre at the point (5, 6)
of radius 5 units.
(c) Find the inverse of the point (5, 10) with respect to a circle with centre at the point
(3, 4) of radius 6 units.
Long Questions
7. (a) Find inversion line segment of AB with points A(1, 2), B(3, 3) with respect to
inversion circle x2 + y2 + 8x + 8y + 24 = 0.
(b) Find the inverse line segment of PQ where P(4, 5) and Q(4, 3) with respect to the
circle x2 + y2 – 4x – 6y = 3, show both of the line segments on the same graph with
the inversion circle.
2. (a) 4 (b) 81 (c) 3 (d) 4 2
(e) 3 (f) 1, (g) 324 (h) 48
(d) (9, 0)
3. (a) 50 , 75 (b) 4 , 8 (c) 64 , 80
13 13 5 5 41 41
4. (a) 12, 6, 4 (b) 32, 16, 8 (c) 10, 5, 2.5
5. (a) 2 , – 5 (b) 12 , 16 (c) 100 , 125
29 29 25 25 41 41
6. (a) (2.8, 2) (b) 60 , 2 (c) 24 , 47
17 17 5 5
7. (a) A'B', A' - 204 , - 196 , B' –24 , 24 (b) P'(6, 7), Q'(10, 3), P'Q'
61 61 7 7
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11.10 Use of Matrices in Transformation
Matrices can be used in most of the elementary transformation.
Let us take a point P(x. y) on the plane.
( ) (Let it be pre-multiplied by ) ( ) ( )0
1
-1 0 then -1 x -x
0 1 0 y =y
Also, P(x, y) y - axis
P'(-x, y)
( )-x
Above two images y and (-x, y) are same image of the point P(x, y).
( )Hence, -1 0 represents a transformation matrix that represents a transformation
0 1
matrix that reflects a given point on Y-axis.
Definition : The matrix which is used to transform a point to get its image is called
transformation matrix.
Translation
Translation by using 2 × 1 matrix
aa
( ) ( )Let us consider T = b , a column vector. Here, T is a 2 × 1 matrix. This matrix b
represents a translation of 'a' unit in the direction of X-axis and 'b' units in the direction of
Y-axis. Y
Let P(x. y) be any point on the plane. P'(x + a, y + b)
a P(x, y) a b
X' O X
( )Let us translate P(x, y) by T = b .
Y'
P T P'
( )x ( a ) ( )x + a
y+ b
= y+b
( )We get that P'(x + a, y + b) is the image of P due to translation T = a .
b
Example: ( )Find the image of P(4, 5) under translation vector -2 .
Solution: 2
( )Given point P(4, 5) can be written as 4
5
( )and -2
translation vector T = 2 which is 2 × 1 matrices.
Now, P T P'
( ) ( ) ( )4 -2 2
5 + 2 =7
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Alternatively,
Let P(x', y') be image of P (4, 5) under translation T we get,
P T P'
( ) ( ) ( ) ( )x' 4 -2 2
y' = 5 + 2 = 7
Hence, P'(2, 7) is the required image of P(4, 5).
Transformation using 2 × 2 matrices
x
[ ]A point (x, y) can be written in the matrix form as y .
[ ]a b
Let it be pre-multiplied by a matrix c d of order 2 × 2.
Then, we get the resulting matrix of order 2 × 1.
[ ] [ ] [ ] [ ]a b
i.e. c d
x ax + by ax + by
y = cx + dy cx + dy is the image of P
[a ]b
under transformation matrix. c d is the transformation matrix
of order 2 × 2.
Example : Let P(3, 4), Q(0, 3), and R(3, 0) the vertices of ∆PQR. Find the image of ∆PQR
Solution:
[ ]0 -1
using 2 × 2 matrix -1 0 .
The vertices of given ∆PQR can be written in the matrix form as,
P QR
[ ]3 0 3
4 30
[ ]Transformation matrix is 0 -1
Now, Image = transformation-1mat0rix × object matrix
P QR [= P' Q' R'
[ ] [ ]0 -1
-1 0 0-4 ]0-3 0+0
3 03 -3+0
4 30 0 + 0 -3 + 0
P' Q' R'
[ ]=
-4 -3 0
-3 0 -3
Hence, the coordinates of image vertices are P'(-4, -3), Q'(-3, 0) and R'(0, -3).
∆ P'Q'R' is the image of ∆PQR.
Reflection using 2 × 2 matrices.
Let P(x, y) be a point and P'(x' y') its image under certain geometrical transformation.
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(a) Reflection on X-axis.
Since we have P(x, y) P'(x, -y)
Which shows that x' = x and y' = - y.
We can write,
x' = 1 . x + 0 . y
y' = 0 . x + (-1) . y
Writing above linear equations in matrix form.
[ ] [ ] [ ]1 0
0 -1
xx
y = -y
[-1 ]0
Hence, 0 1 is the transformation matrix of reflection on y-axis.
(b) Reflection in y - axis
Since P(x, y) y-axis P'(-x, y)
which shows that x' = -x and y' = y ,
We can write x' = x (-1).x + 0.y
y' = y = 0.x + 1.y
Writing above linear equations in matrix form.
[ ] [ ] [ ] [ ]x' -1 0
y' = 0 1
x -x
y= y
[-1 ]0
Hence 0 1 is the transformation matrix of reflection on y-axis.
(c) Reflection on the line y = x.
Since we have,
P(x, y) y= x P'(y, x)
i.e. x' = y and y' = x
x' = y = 0.x + 1.y
y' = x = 1.x + 0.y
Writing these linear equation in the matrix form,
[ ] [ ] [ ] [ ]x' 0 1 y
y' = 1 0 x.
x
y=
[0 ]1
The matrix 1 0 is the transformation matrix of reflection on the line y = x
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(d) Reflection on the line y = -x
Since we have,
P(x, y) y=-x P' (-y, -x)
i.e. x' = -y, y' = -x
x' = -y = 0. x + (-1) y
y' = -x = (-1)x + 0.y
Writing these equations in the matrix form
[ ] [ ] [ ] [ ]x' 0 -1
y' = -1 0
x -y
y = -x .
[ ]0 -1
Hence, the matrix -1 0 describes reflection on the line y = -x.
Rotations Using 2 × 2 Matrices
(a) Rotation of +90° about origin O.
Since we have,
P(x, y) P(-y, x)
i.e. x' = -y and y' = x
x' = 0.x + (-1)y
y' = x = 1.x + 0.y
Writing these equation in matrix form, we write
[ ] [ ] [ ] [ ]x' 0 -1
y' = 1 0
x -y
y= x.
[0 ]-1
Hence, the matrix 1 0 represents a transformation matrix of rotation of +90°
about origin.
(b) Rotation of 180° about origin.
Since we have,
P(x, y) R[0,180°] P'(-x, -y)
i.e. x' = -x and y' = -y
x' = - x = (-1) . x + o.y
y' = -y = o.x + (-1). y
Writing there equations in matrix form, we get,
[ ] [ ] [ ] [ ]x' -1 0
y' = 0 -1
x -x
y = -y .
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[ ]Hence, the matrix
-1 0 represents the rotation of 180° about the origin.
0 -1
(c) Rotation of +270° about origin
Since we have,
P(x, y) R[0,270°] P'(y, -x)
i.e. x' = y and y' = -x
x' = y = 0.x + 1. y
y' = - x = (-1).x + o.y
Writing these equations in the matrix form,
[ ] [ ] [ ] [ ]x' 0 1
y' = -1 0
xy
y = -x .
[ ]Hence, the matrix
01
-1 0 represents rotation of + 270° or -90° about the origin.
Enlargement Using 2 × 2 Matrices
Let O be the centre of enlargement and k the scal factor. Let it be denoted by E[(0, 0), k].
Now, P(x, y) E[(0,0),k] P'(kx, ky)
i.e x' = kx = k.x + 0.y
y' = ky = 0.x + k.y
There equations can be written in the matrix form
[ ] [ ] [ ] [ ]x' k
y' = 0
0 x kx
k= y= ky .
[k ]0
Hence, 0 k represents enlargement matrix with centre O and scale factor k.
Unit Square Y B(1, 1)
The square having vertices 0(0, 0), (1, 0), (1, 1) and (0, 1) is C
called unit square. (0, 1)
In the figure, OABC is a unit square. X' O
Y'
i.e. OA = OC = AB = CB = 1 unit X
In matrix form, unit square is represented by A(1, 0)
[ ]0 1 1 0
00 1 1
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Transformations in terms of matrices can be tabulated as follows:
S.N Transformations Objects Images Equations Matrices
1. Reflection on x-axis (x, y) (x, -y) x'=1.x + 0.y
[ ]1 0
2. Reflection on y-axis (x, y) (-x, y) y'=0.x+(-1)y 0 -1
x'=(-1)x+o.y
3. Reflection on y=x (x, y) (y, x) [ ]-1 0
y'=o.x+1.y 01
4. Reflection on line y=-x (x, y) (-y, -x) x'=o.x+1.y
[ ]0 1
5. Rotation of +90° or (x, y) (-y,x) y'=1.x+o.y 10
+270° about 0 (y, -x) x'=o.x+(-1)y
(-x, -y) [ ]0 -1
6. Rotation of -90° or (x, y) (kx, ky) y'=(-1)x +o.y -1 0
+270° about 0 (x, y) x'=o.x+(-1)y
[ ]0 -1
7. Rotation of 180° about (x, y) y'=1.x+o.y 10
origin x'=o.x+1.y
[ ]0 1
8. Enlargement with centre (x, y) y'=(-1)x+o.y -1 0
origin and scale factor k x'=(-1)x+o.y
[ ]-1 0
9. Identity transformation (x, y) y'=o.x+(-1y) 0 -1
x'=k.x+o.y
[ ]k 0
y'=o.x+k.y 0k
x'=1.x+o.y
[ ]1 0
o.x+1.y 01
Worked out Examples
Example 1. [ ]Find the image of P(2, 3) using matrix which is translated by the vector
Solution: -3
T= 1
Example 2.
Solution: [ ]-3
Here, the translation vector T 1
Translating P by translation vector T, we get,
[ ] [ ] [ ] [ ] [ ]x' 2 -1
-3 2 - 3
y' = 3 + 1 = 3 + 1 = 4 .
[0 ]-1
Which transformation is associated with matrix -1 0 ? Use it to
transform the point (6, 4).
Let P(x, y) be any point on the plane.
[ ] [ ] [ ] [ ]Then -y
0 -1 x 0 + (-y) .= -x .
-1 0 y= -x + 0
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P(x, y) P'(-y, -x)
This shows that P is reflected on line y = -x.
[ ]0 -1
Hence, -1 0 is associated with transformation of reflection on line y = -x.
[ ] [ ] [ ]Again,
0 -1 6 -4
-1 0 4 = -6 .
? (-4, -6) is the image of (6, 4).
[find the image of R(3, 2). ]0 ? Use it to
Example 3. Which transformation is associated with the matrix 3
Solution: 3
0
Let P(x, y) be any point on the plane.
3
[ ] [ ] [ ]Transformation matrix M = 0
0 x 3x
3 y = 3y .
? P(x. y) P'(3x, 3y)
Hence, the given matrix M is associated with enlargement about O with scale
factor 3.
R'
[ ][ ] [ ]Also,
30 39
03 2= 6
Example 4. Write down the 2 × 2 transformation matrix which represents enlargement
Solution: with centre (0, 0) by scale factor 2. Use it find the image of point (4, 3).
Transformation matrix representing enlargement about O and scale factor 2 is
[ ]2 0
M= 0 2 .
Let given point be P(4, 3)
M P P'
[ ] [ ] [ ]2 0
4 8
Now, 0 2 = 3 = 6
? The image of P(4, 3) is P'(8, 6)
Example 5. [4 ]0
To which transformation is the matrix P = 0 4 associated ? What
types of transformation do P2 and P3 represent?
Solution: [4 ]0
Here, P = 0 4 is associated with the enlargement about centre origin
O(0, 0) and scale factor 3.
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40
4
[ ] [ ] [ ]Now, P2 = P × P = 0
40 16 0
0 4 = 0 16
Hence P2 represents an enlargement about centre origin O(o,o) with scale
factor 16.
16 0
[ ][ ][ ]Again. P3 = P2. P = 0 16
40 64 0
04 0 64 .
Hence P3 represents an enlargement matrix about centre origin O(0, 0) with scale
factor 64.
Example 6. The vertices of a unit square are O(0, 0), A(1, 0), B(1, 1) and C(0, 1), The
Solution: square OABC is reflected on the line y = x. Find the coordinates of image
using 2 × 2 matrix.
Given unit square OABC can be written in the matrix form.
OA B C
[ ]0 1 1 0
00 11
01
0
[ ]2 × 2 matrix of reflection on line y = x is M = 1
OA B C O' A' B' C'
[ ][ ] [ ]Now, 1 0 0 0 1 1 = 0 1 1 0 .
01 0 1 1 0 00 11
Hence, the coordinates of image of unit square are O(0, 0), A(0, 1), B(1, 1), and
C'(1, 0).
Example 7. Find the image of ∆ABC with vertices A(2, 3), B(-1, 2), and C(2, -3) using
[ ] [ ]the matrix
0 -1 0 -1
-1 0 . Which transformation is the matrix -1 0
associated with ?
Solution: Writing the coordinates of vertices A(2, 3), B(-1, 2) and C(2, -3) of ∆ABC in the
AB C
[ ]2 -1 2
matrix form , 3 2 -3
0 -1
[ ]Given transformation matrix is (M) = -1 0
AB C A' B' C'
[ ] [ ] [0 -1
-3 -2 ]3
Now, -1 0
2 -1 2 -2
3 2 -3 = -2 1
Hence, the coordinates of image of the ∆ABC are
A'(-3, -2), B'(-2, 1) and C'(3, -2).
[ ]The matrix
0 -1 is associated with reflection on the line y = -x.
-1 0
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Example 8. [Transform a unit square under the matrix 3 ]2
Solution: coordinates of the image. 1
1 and write the
Let the vertices be O(0, 0), A(1, 0), B(1, 1) and C(0, 1) of a unit square.
OA B C
01 10
[ ]Writing the given unit square in the matrix form 0 0 1 1
[ ]3 2
Transformation matrix, M = 1 1 .
Now, M × object OABC = Image O'A'B'C'.
OA B C O' A' B' C'
[ ][ ] [ ]i.e., 1 1 0 0 1 1 = 0 1 2 1
32 0 1 1 0 03 52
Hence the coordinates of image vertices are O'(0, 0). A(3, 1), B'(5, 2), C(2, 1)
Example 9. ∆PQR whose vertices are P(3, 6), Q(4, 2), and R(1, 1) maps onto ∆P'Q'R'
Solution: with the vertices P'(-6, 3), Q'(-2, 4), and R'(-2, 2). Which is the single
transformation for this mapping? Also find a 2 × 2 matrix that represents
this transformation.
Here, ∆PQR with vertices P(3, 6), Q(4, 2), and R(1, 1) is mapped onto ∆P'Q'R'
with the vertices P'(-6, 3), Q(-2, 4), and R(-1, 1)
Here, P(3, 6) P'(-6, 3)
i.e. (x, y) (-y, x)
Hence, the point (x, y) is transformed to (-y, x) which represents the rotation
of +90° about the origin.
[a ]b
LetM = c d be the required 2×2 transformation matrix which transform
∆PQR to ∆P'Q'R' .
Then,
AB C P' Q' R'
[ ] [ ] [ ]a b
2 -1 2 -6 -2 -1
c d . 3 2 -3 = 3 4 1 .
[ ] [ ]or, 3c + 6d 4c + 2d c + d = 3 4 1 .
3a + 6b 4a + 2b a + b -6 -2 -1
Equating the corresponding elements, we get,
3a + 6b = -6 or, a + 2b = -2........ (i)
4a + 2b = -2 or, 2a + b = -1........ (ii)
a + b = -1 ........ (iii)
3c + 6d = 3 or, c + 2d = 1 ....... (iv)
4c + 2d = 4 or, 2c + d = 2....... (v)
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c + d = 1 ......... (vi)
Solving equation (i) and (iii), we get,
a + 2b = -2
a + b = -1
-- +
b=-1
Put the value of 'a' in equation (iii), we get,
a=0
Again solving equations (iv) and (vi), we get,
c + 2d = 1
c+ d = -1
-- +
d=0
Put the value of d in equation (vi), we get, c = 1
[ ]0 -1
? M = 1 0 which is the required matrix.
[Example 10. Find the 2 × 2 matrix which transforms the unit square 0 1 1 ]0
[ ]to a parallelogram03 4 1 0 0 1
0 1 3 2 1
.
Solution: [Let M = a ]b be the required matrix. Then by question
c
d
[ ][ ] [a b 0 1 1 0 = 0 3 4 ]1
cd 0 0 1 1 0 1 3
[ ] [ ]or, 2
0 a a+b b = 0 3 4 1
0 c c+d d 0 1 3 2
Equating the corresponding elements of equal matrices, we get,
a = 3, b = 1, d = 2, c = 1
[ ].Required transformation matrix is M =3 1
1 2
Example 11. Prove by matrix method that the reflection in x-axis followed by rotation
about (0, 0) through 180° is equivalent to the reflection on y-axis.
Solution: The 2 × 2 transformation matrix representing the reflection on x-axis
0 1 -1 0
-1 0 -1
[ ] [ ]0
is and rotation about O(0, 0) through 180° is
Now, the reflection followed by rotation about O(0, 0) through 180° is given
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by
[ ] [ ] [ ]-1 01 0 = -1 0 ............................. (i)
0 -1 0 -1 0 1
The 2 × 2 transformation matrix representing reflection in y-axis is
[ ]-10 .............. (ii)
0 1
From (i) and (ii), we conclude that the reflection on x-axis followed by rotation
about (0, 0) through 180° is equivalent to the reflection on y-axis.
Example 12. A unit square is rotated about origin through 180°. Find the image of unit
square using 2 × 2 matrix. Plot the graphs of unit square and its image on
the same graph paper.
Solution: Let O(0, 0). A(1, 0). B(1, 1) and C(0, 1) be the vertices of unit square OABC.
OA B C
[ ]Theunitsquarecanbeexpressedinthematrixformas.
0 1 1 0
0 0 1 1
-1 ]0
[The 2 × 2 matrix of rotation about origin through 180° is 0 -1 .
C'
OA B C O' A' B'
[ ][ ] [-1 0 0 1 1 0 ]0
0 -1 -1
Now, 0 -1 0 0 1 1 = 0 0 -1 -1
? The required image vertices of unit square are O'(0, 0), A'(-1, 0), B'(-1.
-1). C'(0, -1).
Graphs of unit square OABC and its image O'A'B'C' are drawn in the graph
given below.
Scale 20 Y
lines - 1 unit
B'
C(0, 1) B(1, 1)
X' A' X
O
A(1, 0)
C'
Y'
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Exercise 11.6
Very short Questions
1. Let P(x, y) be any point on the plane and P(x', y') be the image of P. Write the transformation
matrices under the following cases:
(a) (x', y') = (-y, - x) (b) (x', y') = (-x, -y) (c) (x', y') =(-x, -y)
(d) (x', y') = (-x, y) (e) (x', y') = (y, x) (f) (x' , y') =(kx, ky)
(g) (x', y') = (x + a, y + b) (h) (x', y') = (3x, 3y)
2. Write the type of transformation represented by each of the following transformation
matrices:
( )1 0 ( )1 0 ( )0 1
(a) 0 1 (b) 0 -1 (c) 1 0
( )0 -1 ( )0 -1 ( )0 1
(d) -1 0 (e) 1 0 (f) -1 0
( )-1 0 ( )2 0 (4 )0
(g) 0 -1 (h) d 2 (i) 0 4
Find the image of P(4, 5) using above matrices in each care.
Short Questions
3. Write down 2 × 2 transformation matrices in each of the following transformation and
using the matrix find image of given point.
(a) Reflection on y-axis, A(3, 4)
(b) Reflection on x-axis. B(6, 3)
(c) Reflection on line y = x , C(-4, -2)
(d) Reflection on line x + y = 0, D(2, 3)
(e) Rotation about origin through, +90°, E(-4, -4)
(f) Rotation about origin through 180°, F(6, 7)
(g) Enlargement with centre at the origin by scale factor 3, G(4, 7)
[ ]4. (a) If a matrix
value of a. a0
3 2 maps the point (3, 4) on to the point (15, 17), find the
(b) If a point (x, y) is transformed into (y, x) by a 2 × 2 matrix, find the 2 × 2
transformation matrix.
(c) Find a 2 × 2 matrix which transforms point (8, -4) to P'(4, 8).
2
[ ]5. (a) Find the image of A(6, 7), under the translation matrix 4 followed by
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[ ]3
translation 2 .
34
[ ] [ ](b) Find the image of M(2. 3) under translation 2 followed by translation 2 .
2 ]0
[6. (a) To which transformation is the matrix M= 0 2 associated with ? What
types of transformation do M2 and M3 represent ?
[1 ]0
(b) What does the matrix 0 -1 represent? Find the image of P (4, 5) using the
matrix.
Long Questions
7. (a) P(2, 1), Q(5, 1), R(5, 4) and S(2, 4) are the vertices of a square PQRS. The square
[1 ]2
is transformed by matrix 1 -2 into a parallelogram. Find the vertices the
parallelogram.
(b A triangle with vertices P(2, 1), Q(5, 3), and R(6, 7) is transformed by the matrix
[ ]0 -1
-1 0 . Find the image of the triangle.
(c) A square with vertices A(1, 1), B(6, 1), C(6, 6), and D(1, 6) is transformed by the
[transformation matrix 3 ]0
0
3 . Find the image of the square ABCD.
(d) A quadrilateral PQRS with vertices P(2, 1) Q(5, 2), R(6, 5) and S(3, 5) is transformed
[ ]-1 0
by matrix 0 -1 find the image of the quadrilateral PQRS.
8. (a) A line PQ having P(5, 1), Q(8, 6) map's onto P'Q' having P'(-5, 1), Q(-8, 6) so that
the image P'Q' is formed. What is the single transformation for the mapping ? Also
find the 2 × 2 matrix.
(b) A square ABCD with vertices A(2, 0), B(5, 1) C(4, 4) and D(1, 3) is mapped onto
parallelogram A'B'C'D' by a 2 × 2 matrix so that the vertices of the parallelogram
are A'(2, 2), B'(7,3) C'(12, - 4) and D'(7, -5). Find the 2 × 2 matrix.
(c) ∆ABC with the vertices A(2, 7), B(2, 9), and C(6, 7) is mapped onto ∆A'B'C' whose
vertices are A'(7, 2), B(9, 2) and C'(7, 6). Which is the single transformation for the
mapping? Also find the 2 × 2 matrix of transformation ?
(d) ∆PQR with vertices P(5, 1), Q(12, 4) and R(4, 5) maps onto the ∆P'Q'R' with the
vertices P'(-5, -1) Q'(-12, -4) and R'(-4, -5). Which is the single transformation for
this mapping ? Also find the 2 × 2 matrix that represents the transformation.
9. (a) A unit square having vertices A(o. o), B(1, 0), C(1, 1) and D(0, 1) is mapped to the
parallelogram A'B'C'D' by a 2 × 2 matrix so that the vertices of parallelogram are
A'(o,o), B'(3, 0), C(4, 1) and D'(1, 1). Find the 2 × 2 transformation matrix.
(b) Find the 2 × 2 matrix which the unit square is transformed to parallelogram
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( )0 3 4 1 .
0 1 3 2
[ ](c) 0 1 1 0
Find the 2 × 2 matrix which transforms the unit square 0 0 1 1
[ ]onto a parallelogram 0 4 6 2 .
0 1 3 2
[ ](d) A rectangle
02 20 is mapped onto a rectangle
00 -1 -1
[ ] [ ]0
0 2 2 0 a b , find the matrix of
0 1 1 by the 2 × 2 matrix c d
transformation.
(e) A unit square OMNP having vertices O(o, o), M(1, 0), N(1, 1), and P(0, 1) is
transformed to unit, square O'M'N'P' through reflection y = -x. Write the vertices
of the image unit square so formed. Draw the graphs of both of the units square on
the same graph paper.
10. (a) Verify by the matrix method that the reflection on the line y = x followed by the
reflection on the y-axis is equivalent to rotation about origin through +90°.
(b) Verify by the matrix method that a reflection in the line X - axis followed by the
Y - axis is the rotation about the origin through 180°.
(c) Prove by the matrix method that the reflection on the line y = x followed by the
rotation about the origin +90° is the reflection on y-axis.
3. (a) A'(-3, 4) (b) B'(6, -3) (c) C'(-2, -4) (d) D'(-3, -2)
(e) E'(4, -4) (f) F'(-6, -7) (g) G'(12, 21)
4. (a) 5
5. (a) A'(11, 13) (b) 01 (c) 0 -1
10 10
(b) M'(9, 7)
6. (a) E[(0, 0), 2], M2 represents E[(0, 0), 4], M3 represents E[(0, 0), 8]
(b) Reflection in X-axis, P'(4, -5)
7. (a) P'(4, 0), Q'(7 3), R'(13, -3), S'(10, -5) (b) A'(-1, -2), Q'(-3, -5), R'(-7, -6)
(c) A'(3, 3), B'(18, 3), C'(18, 18), D'(3, 18) (d) P'(-2, -1), Q'(-5, -2) R'(-6, -5), S'(-3, -5)
8. (a) -1 0 (b) 12 (c) reflection on y = x, 0 1
0 1 1 -2 1 0
(d) Rotation of 180° about origin, -1 0
0 -1
9. (a) 3 1 (b) 31 (c) 42 (d) 10
0 1 12 12 0 -1
(e) O'(0, 0), M'(0, -1), N(-1, -1)
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12Statistics
12.0 Measure of Dispersion
Introduction
Measures of central tendency gives us an idea of the concentration of observations about the
central part of the distribution. It tells nothing about how each variate value of a set of data is
scattered from an average value of those items. Two or more two data may have equal mean,
median or mode but with more scattered than the other. It may be clear from the following
two set of data.
Averages X Md Mo
Set I 63 64 65 66 66 67 68 69 66 66 66
Set II 64 65 66 66 66 66 67 68 66 66 66
The above two data sets have the same means, the same medians and same mode, all equal to
66. They have the same number of observations, n = 8. But these two data set are different.
What is the main difference between them ?
The two data sets have the same measure of central tendency (as measured by any of the
three measures of central tendency - mean, median and model) but they have a different
variability. In particular, we can see that data set I is more scattered than data set II. The
values of data set I are more spread out: they lie farther away from their mean than do those
data set II.
Hence, the certain measures are evolved which reflect on the scattering of values of
numerical terms are known as measures of dispersion. In statistics, the term dispersion is
used commonly to mean scatteredness, variation, diversity, deviation, fluctuations, spread,
heterogeneity, etc.
Dispersion may be defined as the degree of scatteredness or variation of a variable about a
central value. Less the value of dispersion more data will be representative. Dispersion of the
data are measured in terms of the following measures of dispersion.
1. Range 2. Quartile Devication
3. Mean Deviation 4. Standard Deviation
In this chapter we study about quartile deviation, mean deviation, and standard deviation
for continuous series of data.
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12.1 Quartile Deviation (Q.D)
Quartile deviation is a measure of dispersion based on the upper quartile (Q3) and the lower
quartile (Q1) of a set of data. The difference between the upper quartile (Q3) and the lower
quartile (Q1) is called interquartile range.
Interquartile range = Q3 - Q1
Half of the interquartile range is called semi-interquartile range or quartile deviation. It is
denoted Q.D. and is given by the formula.
Quartile Deviation (Q.D.) = 1 (Q3 - Q1)
2
It is an absolute measure of dispersion. It has the same unit as that of original data.
For comparative study of two or more data sets, we calculate coefficient of quartile deviation.
It is the relative measure of dispersion based the quartiles Q3 and Q1
Coefficient of Quartile Deviation = Q3 - Q1
Q3 + Q1
The coefficient of Q.D. is unitless. It is used to compare variability of two or more data
sets. The data having less value of coefficient of quartile deviation is less variable or less
heterogeneous or more uniform or more stable than other and vice-versa.
Calculation of Quartile Deviation for continuous Series.
For continuous data in the form of frequency distribution, the following formulas are used.
For lower quartile or the first quartile (Q1)
N
Find the value of 4 for the first quartile class.
N - c.f
4
Q1 = L + ×i
f
Where, (i) L is the lower limit of the class in which the first quartile lies.
(ii) f is the frequency of the first quartile class.
(iii) c.f. is the cumulative frequency of the class preceding the first quartile class.
(iv) i is the length of class interval (or class - size).
For the upper quartile or the third quartile (Q3)
Find the value of 3N to find the third quartile class.
4
3N
4 - c.f
Q3 = L + f ×i
Where, (i) L is the lower limit of the class in which the third class lies.
(ii) f is the frequency of the third quartile class.
(iii) c.f. is the cumulative frequency preceding the third quartile class.
(iv) i is the length of class interval (or class size).
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vedanta Excel In Opt. Mathematics - Book 10
Worked Out Examples
Example 1. The quartile deviation of a continuous data is 20. If the first quartile is 45,
Solution: find the third quartile.
Example 2. Given, Quartile Deviation (Q.D.) = 20
Solution:
The first quartile (Q1) = 45
Example 3.
Now, Q.D. = 1 (Q3 - Q1)
2
1
or, 20 = 2 (Q3 - Q1)
or, 20 = 1 (Q3 - 45) or, 40 = Q3 - 45
2
? Q3 = 85
The coefficient of quartile deviation of a grouped data is 0.25. If the upper
quartile is 75, find the lower quartile.
Coefficient of Q.D. = 0.25
The upper quartile (Q3) = 75
Now, coefficient of Q.D. = Q3 - Q1
Q3 + Q1
5041Q.215===27727755555++-- QQQQ1111
or, or, 75 + Q1 = 300 - 4Q1
? Q1 = 45
or,
or,
Calculate the quartile deviation and its coefficient from the following data :
Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of students 1 18 25 28 30 33 22 15 22
Solution: To calculate quartile deviation and its coefficient.
Marks obtained No. of students c. f.
0-10 11 11
10-20 18 29
20-30 25 54
30-40 28 82
40-50 30 112
50-60 33 145
60-70 22 167
70-80 15 182
80-90 22 204
N = 204
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 349
vedanta Excel In Opt. Mathematics - Book 10
To find Q1 = N = 204 = 51
4 4
c.f., just greater 51 is 54. So, Q1 lies in the class (20 - 30). i.e. (20-30) is the first
quartile class.
L = 20, c.f = 29, f = 25, i = 10
N - c.f
4
Now, Q1 =L+ ×i
f
51 - 29
= 20 + 25 × 10 = 20 + 8.8 = 28. 8 marks.
To find Q3 , 3N = 3 × 204 = 153
4 4
c.f. just greater 153 is 167.
Q3 lies in the class (60 - 70) i.e. (60 - 70) is the third quartile class.
L = 60, c.f. = 145, f = 22, i = 10
3N - c.f
4
Now, Q3 =L+ ×i
f
Dev=iat6io0n+(Q.1D5)3=2-21214(5Q3×- Q110)
= 60 + 3.64 = 63.64 marks
Quartile = 1 (63.64 - 28.8) =17.42
2
Q3 - Q1 63.64 - 28.8
coefficient Quartile Deviation (Q.D) = 2 = 63.64 + 28.8 = 0.38
Example 4. Calculate the quartile deviation and its coefficient.
x less than 10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80 and above
f3 4684783 5
Solution: The given data is open ended. We can calculate deviation.
x f cf
Less than 10 3 3
4 7
10-20 6 13
20-30 8 21
30-40 4 25
40-50 7 32
50-60 8 40
60-70 3 43
70-80 5 48
80 and above N = 48
To find the first quartile (Q1)
N = 48 = 12
4 4
350 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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Vedanta Opt. Maths Book 10 Final (2078)
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